y=(x²+2x+2)/(x+1)
=[(x+1)²+1]/(x+1)
=x+1+[1/(x+1)]
≥2
當(dāng)且僅當(dāng)x+1=1/(x+1)時(shí)即x=0或x=-2時(shí)取到等號(hào)2是哪來(lái)的啊？答案上還有一個(gè)解是≤-2呢根據(jù)基本不等式a+b≥2ab，這里的a是x+1，b是1/(x+1)用判別式法解這道題是（-無(wú)窮，-2】∪【2，+無(wú)窮），用分離常數(shù)法怎么解出≤-2這個(gè)解?。坎缓靡馑冀忮e(cuò)了。
y=(x²+2x+2)/(x+1)
=[(x+1)²+1]/(x+1)
=x+1+[1/(x+1)]
當(dāng)x+1≥0即x≥-1時(shí)
根據(jù)基本不等式a+b≥2√ab
y≥2，當(dāng)且僅當(dāng)x+1=1/(x+1)時(shí)即x=0或x=-2(舍)時(shí)取到等號(hào)
當(dāng)x+1＜0即x＜-1時(shí)
原式=-[(-x-1)+1/(-x-1)]
根據(jù)基本不等式a+b≥2√ab
y≤-2，當(dāng)且僅當(dāng)x+1=1/(x+1)時(shí)即x=0(舍)或x=-2時(shí)取到等號(hào)